The idea of a mathematical proof is that without dependence of a specific input, if certain properties are true, then the conclusion is also true.
In this case, if the composition is injective then it has to be that the "outer" function is injective, and that if the functions are surjective then their composition is as such.
Showing a specific case is a valid method for disproving a claim, as it shows that at a certain time the properties hold but the conclusion is false.
For example, consider $A=\{0\}$ and $B=\{1,2\}$ and $C=\{0\}$ and $f(0)=1$ and $g(1)=0, g(2)=0$.
The composition $g\circ f$ is a function from $A$ to $C$, since $A$ is a singleton every function is injective. However $g$ is clearly not injective.
This is a counterexample to the claim, thus disproving it.
The proof for the second one, if both functions are surjective then so is their composition, let us check this.
Let $A,B,C$ any sets, and $f,g$ functions as required. To show that $g\circ f$ is surjective we want to show that every element of $C$ is in the range of $g\circ f$. Assume $c\in C$ then there is some element $b\in B$ such that $g(b)=c$. Since $f$ is surjective we have some $a\in A$ such that $f(a)=b$. It follows that $g\circ f (a) = c$ and therefore the composition of surjective functions is surjective.
Now, consider this proof. There was nothing to limit ourself. I did not require anything from $A,B,C$ other than them being sets, and I did not require anything from $f,g$ other than them being functions from/onto the required set in the premise of the claim. Furthermore, I did not check the surjective-ness of the composition by choosing a certain element. I picked and arbitrary element, from a set which was also arbitrary. Using this sort of argument assures us that the property is not dependent on any of the characteristics of the set (for example, some things are only true on finite, or infinite, or other certain sets). This allows us to have a very general theorem. Whenever we have three sets, and two functions which are surjective between them, then the composition is also surjective!
What I know about your nice question is as follows. I suppose that $f:A\to B$ and $g:B\to C$.
If $g\circ f$ is 1-1 then $f$ is 1-1.
If $g\circ f$ is onto then $g$ is onto.
so if $g\circ f$ is bijective then $f$ is 1-1 and $g$ is onto but, the converse is not true.
$f(x)=f(x')\Longrightarrow g(f(x))=g(f(x'))\Longrightarrow (g\circ f)(x)=(g\circ f)(x')\Longrightarrow x=x'$
if $c\in C$ then $\exists a (a\in A \wedge (g\circ f)(a)=c)\Longrightarrow\exists a(a\in A \wedge (g\big(f(a)\big)=c)$ so if we take $b=f(a)\in B$ then $g(b)=c$.
Now take $A=\{1,2\}, B=\{3,4,5\}, C=\{6,7\}$ and $$f:A\to B\\f(1)=3,f(2)=4$$ $$g:B\to C\\ g(3)=g(4)=6,g(5)=7$$ $f$ is 1-1 and $g$ is onto but $$(g\circ f)(1)=g(3), (g\circ f)(2)=6$$ which means that $g\circ f$ is not a bijective map.
Best Answer
"if $f$ injective and $g$ surjective can $g\circ f$ be both injective and surjective?"
Sure, e.g. let $f:\{0\}\rightarrow\mathbb R$ and $g:\mathbb R\rightarrow\{0\}$.
But not always, e.g. let $g:\{0,1,2\}\rightarrow\{0,1\}$ be prescribed by $0\mapsto0$, $1\mapsto1$, $2\mapsto1$ and let $f:\{0,1\}\rightarrow\{0,1,2\}$ be prescribed by $0\mapsto1$, $1\mapsto2$.
Then $f$ is injective and $g$ is surjective.
However $g\circ f:\{0,1\}\rightarrow\{0,1\}$ is constant (and prescribed by $0\mapsto1$ and $1\mapsto1$).
It is evidently not injective and is not surjective.